Question

5.
A ball is thrown at an angle of 30 degree to the horizontal. Ii falls on the ground at a distance of 90 m.
If the ball is thrown with the same initial speed at an angle 30° to the vertical, it will fall on the
ground at a distance of
(A)120 m
(B)27m
(C)90 m
(D)30 m​

Answer 1

GivEn :

• At 30° ball is thrown upwards .

• Distance on which the ball falls = 90 m

• The ball is thrown with the same initial velocity = 30° to the vertical.

To find :

• The distance in which the ball falls on the ground = ?

SoluTion :

To calculate the distance, we should know the appropriate formula for it!

✩ Distance covered by the ball is given as,

$\star \: { \boxed { \sf{ \purple{\: x = \dfrac{ {v}^{2} \sin 2 \theta }{g} }}}}$

where,

• v = velocity at which the ball is thrown

• g = gravitational force

Now,

According to the question, ball is thrown at 30° angle to the horizontal, so,

• x = 90 m [Given]

• θ = 30°

Substituting the values,

$\sf \dashrightarrow \: \: \: x = \dfrac{ {v}^{2} \sin 2 \theta }{g}$

$\sf \dashrightarrow \: \: \: 90= \dfrac{ {v}^{2} \sin 2 \times 30 }{10}$

$\sf \dashrightarrow \: \: \: 900= {v}^{2} \sin 60$

$\sf \dashrightarrow \: \: \: 900= {v}^{2} \dfrac{ \sqrt{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \bigg( \sin {60}^{ \circ} = \dfrac{ \sqrt{3} }{2} \bigg)}$

$\sf \dashrightarrow \: \: \: 1800= {v}^{2} \sqrt{3}$

$\sf \dashrightarrow \: \: \: {v}^{2} = \dfrac{1800}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: - eq(1).$

Now,

✩ Distance covered by the ball on the ground :

$\sf : \implies \: \: \: x = \dfrac{ {v}^{2} \sin 2 \theta }{g}$

$\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \sin 2 \times 30 }{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg(from \: eq(1). \bigg)$

$\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \sin 60 }{10}$

$\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \frac{ \sqrt{3 } }{2} }{10}$

$\sf : \implies \: \: \: x = \dfrac{ 900 }{10}$

$\sf : \implies \: \: \: { \underline{ \boxed{ \sf{ \red{x = 90 \: m}}}}} \: \bigstar$

Therefore, the ball will fall on the ground at distance of 90 m.

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Answer 2

Explanation:

GivEn :

At 30° ball is thrown upwards .

Distance on which the ball falls = 90 m

The ball is thrown with the same initial velocity = 30° to the vertical.

\begin{gathered} \\ \\ \end{gathered}

To find :

The distance in which the ball falls on the ground = ?

\begin{gathered} \\ \\ \end{gathered}

SoluTion :

To calculate the distance, we should know the appropriate formula for it!

✩ Distance covered by the ball is given as,

\star \: { \boxed { \sf{ \purple{\: x = \dfrac{ {v}^{2} \sin 2 \theta }{g} }}}}⋆

x=

g

v

2

sin2θ

where,

v = velocity at which the ball is thrown

g = gravitational force

Now,

According to the question, ball is thrown at 30° angle to the horizontal, so,

x = 90 m [Given]

θ = 30°

Substituting the values,

\begin{gathered}\sf \dashrightarrow \: \: \: x = \dfrac{ {v}^{2} \sin 2 \theta }{g} \\ \\ \\ \\ \end{gathered}

⇢x=

g

v

2

sin2θ

\begin{gathered}\sf \dashrightarrow \: \: \: 90= \dfrac{ {v}^{2} \sin 2 \times 30 }{10} \\ \\ \\ \\ \end{gathered}

⇢90=

10

v

2

sin2×30

\begin{gathered}\sf \dashrightarrow \: \: \: 900= {v}^{2} \sin 60 \\ \\ \\ \\ \end{gathered}

⇢900=v

2

sin60

\begin{gathered}\sf \dashrightarrow \: \: \: 900= {v}^{2} \dfrac{ \sqrt{3} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \bigg( \sin {60}^{ \circ} = \dfrac{ \sqrt{3} }{2} \bigg)} \\ \\ \\ \\ \end{gathered}

⇢900=v

2

2

3

(sin60

∘

=

2

3

)

\begin{gathered}\sf \dashrightarrow \: \: \: 1800= {v}^{2} \sqrt{3} \\ \\ \\ \\ \end{gathered}

⇢1800=v

2

3

\begin{gathered}\sf \dashrightarrow \: \: \: {v}^{2} = \dfrac{1800}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: - eq(1). \\ \\ \\ \\ \end{gathered}

⇢v

2

=

3

1800

−eq(1).

Now,

✩ Distance covered by the ball on the ground :

\begin{gathered}\sf : \implies \: \: \: x = \dfrac{ {v}^{2} \sin 2 \theta }{g} \\ \\ \\ \\ \end{gathered}

:⟹x=

g

v

2

sin2θ

\begin{gathered}\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \sin 2 \times 30 }{10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg(from \: eq(1). \bigg) \\ \\ \\ \\ \end{gathered}

:⟹x=

10

3

1800

×sin2×30

(fromeq(1).)

\begin{gathered}\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \sin 60 }{10} \\ \\ \\ \\ \end{gathered}

:⟹x=

10

3

1800

×sin60

\begin{gathered}\sf : \implies \: \: \: x = \dfrac{ { \frac{1800}{ \sqrt{3} } } \times \frac{ \sqrt{3 } }{2} }{10} \\ \\ \\ \\ \end{gathered}

:⟹x=

10

3

1800

×

2

3

\begin{gathered}\sf : \implies \: \: \: x = \dfrac{ 900 }{10} \\ \\ \\ \\ \end{gathered}

:⟹x=

10

900

\begin{gathered}\sf : \implies \: \: \: { \underline{ \boxed{ \sf{ \red{x = 90 \: m}}}}} \: \bigstar \\ \\ \end{gathered}

:⟹

x=90m

★

Therefore, the ball will fall on the ground at distance of 90 m.

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